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If $\;f(x)\;$ is continuous and $\;f(\large\frac{9}{2}) = \large\frac{2}{9}\;$ , then $\;\lim_{x}^{0} f(\large\frac{1-cos 3x}{x^{2}})\;$ is equal to :

$(a)\;\large\frac{9}{2} \qquad(b)\;\large\frac{2}{9} \qquad(c)\;0\qquad(d)\;\large\frac{8}{9} $

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Given limit can be represented as  ,   f( 9/2 * [sin^2(3x/2)]/[3x/2]^2 ) which as x tends to 0 is  , f(9/2) = 2/9
answered Apr 2, 2015 by Limitsofawesomeness
 

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